Continuity & the Intermediate Value Theorem

You’ll be able to

State the three-part definition of continuity at a point
Classify discontinuities as removable, jump, or infinite
Apply the Intermediate Value Theorem to guarantee a solution on an interval

Continuity in three conditions

A function f is continuous at x = a when three things all hold: (1) f(a) is defined, (2) lim(x→a) f(x) exists, and (3) those two agree, lim(x→a) f(x) = f(a). Informally, you can draw the graph through x = a without lifting your pencil. If any one condition fails, f is discontinuous at a — and identifying which condition fails tells you what kind of break it is.

Three flavors of discontinuity

A removable discontinuity (a hole) occurs when the limit exists but does not equal f(a) — you could "remove" it by redefining one point. A jump discontinuity occurs when the left and right limits both exist but disagree, so the graph jumps. An infinite discontinuity occurs at a vertical asymptote, where the function blows up. Removable is the only kind you can patch with a single point.

The Intermediate Value Theorem

The Intermediate Value Theorem (IVT) says: if f is continuous on the closed interval [a, b], then f takes on every value between f(a) and f(b) somewhere on that interval. Practically, if f(a) and f(b) have opposite signs, then f must equal 0 at least once in between — a guaranteed root. Continuity on the whole closed interval is the non-negotiable hypothesis; without it the guarantee collapses.

Continuity at a point

f continuous at a ⟺ f(a) defined AND lim(x→a) f(x) exists AND lim(x→a) f(x) = f(a)

All three conditions. The IVT requires this at every point of a closed interval [a, b].

Worked example

Show that f(x) = x³ + x − 1 has a root somewhere in the interval [0, 1].

1.f is a polynomial, so it is continuous everywhere, including on the closed interval [0, 1] — the IVT hypothesis is met.
2.Evaluate the endpoints: f(0) = 0 + 0 − 1 = −1.
3.And f(1) = 1 + 1 − 1 = 1.
4.The values −1 and 1 have opposite signs, so 0 lies between f(0) and f(1).
5.By the IVT, there is some c in (0, 1) with f(c) = 0.

Answer: Because f is continuous on [0, 1] and f(0) = −1 < 0 < 1 = f(1), the IVT guarantees a c in (0, 1) where f(c) = 0.

Watch out

The IVT guarantees a value exists; it never tells you where or how many. And it requires continuity on the entire closed interval — if the function has an asymptote inside [a, b], the sign change proves nothing.

Checkpoint

A function has lim(x→2) f(x) = 5 but f(2) = 1. What type of discontinuity is at x = 2?

Checkpoint

For which reason does the IVT guarantee that g(x) = cos(x) − x has a solution to cos(x) = x on [0, 1]? (g(0) = 1, g(1) ≈ −0.46)

On the exam

A full-credit IVT argument has three moving parts: state that f is continuous on [a, b], compute both endpoint values to show the target lies between them, then conclude "by the IVT" a c exists. Skipping the continuity sentence loses the point even when the arithmetic is right.

Answer the 2 checkpoints as you read.